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A rectangular shamba is x meters wide. Its length is 2 meters longer than its width. a) Write an expression for the length. b) Write an expression for the area.

Mathematics

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Answers

2 parts
a

x+2

b

x(x+2)

400

10−3.206610^{-3.2066}

Step-by-step solution

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Here are the solutions to the problems:

3. Rectangular Shamba
Let the width of the rectangular shamba be xx meters.

Part a

The length

The length is 2 meters longer than the width.

L=x+2L = x + 2

The expression for the length is x+2\boxed{x+2}.

Part bx(x+2)

The area

The area of a rectangle is given by length ×\times width.

A=L×xA = L \times x

Substitute the expression for LL:

A=(x+2)xA = (x+2)x

The expression for the area is x(x+2)\boxed{x(x+2)}.
(Note: The problem states the area is 48 m², so x(x+2)=48x(x+2) = 48, but the question asks for an expression for the area, not to solve for xx.)

4. Farmer distributing rice

  1. Identify the total amount of rice and the capacity of each bag.

    Total rice = 5000 kg
    Rice per bag = 12.5 kg

  2. Determine the number of bags needed.

    The number of bags is the total rice divided by the rice per bag.

    Number of bags=Total riceRice per bag\text{Number of bags} = \frac{\text{Total rice}}{\text{Rice per bag}} Number of bags=500012.5\text{Number of bags} = \frac{5000}{12.5}

  3. Use the reciprocal of 12.5 as requested.

    The reciprocal of 12.5 is 112.5\frac{1}{12.5}.

    Number of bags=5000×112.5\text{Number of bags} = 5000 \times \frac{1}{12.5}

  4. Calculate the reciprocal of 12.5.

    112.5=1252=225=0.08\frac{1}{12.5} = \frac{1}{\frac{25}{2}} = \frac{2}{25} = 0.08

  5. Multiply the total rice by the reciprocal.

    Number of bags=5000×0.08\text{Number of bags} = 5000 \times 0.08 Number of bags=400\text{Number of bags} = 400

    The total number of bags needed is 400\boxed{400}.

    5. Medicine concentration

  6. Write the given concentration.

    Concentration C=0.0006214C = 0.0006214 g/mL.

  7. Express the concentration in scientific notation.

    Move the decimal point 4 places to the right to get a number between 1 and 10.

    C=6.214×10−4 g/mLC = 6.214 \times 10^{-4} \text{ g/mL}

  8. Express the concentration in the form 10n10^n.

    We need to find nn such that 10n=0.000621410^n = 0.0006214.
    This means n=log⁡10(0.0006214)n = \log_{10}(0.0006214).
    Using the scientific notation:

    n=log⁡10(6.214×10−4)n = \log_{10}(6.214 \times 10^{-4})

    Apply the logarithm property log⁡(AB)=log⁡A+log⁡B\log(AB) = \log A + \log B:

    n=log⁡10(6.214)+log⁡10(10−4)n = \log_{10}(6.214) + \log_{10}(10^{-4})

    Apply the logarithm property log⁡(AB)=Blog⁡A\log(A^B) = B \log A:

    n=log⁡10(6.214)−4n = \log_{10}(6.214) - 4

  9. Find log⁡10(6.214)\log_{10}(6.214) using logarithm tables.

    From logarithm tables:
    log⁡10(6.21)=0.7931\log_{10}(6.21) = 0.7931
    The mean difference for 4 (in the row for 62) is 3.
    So, log⁡10(6.214)=0.7931+0.0003=0.7934\log_{10}(6.214) = 0.7931 + 0.0003 = 0.7934.

  10. Substitute the value back into the equation for nn.

    n=0.7934−4n = 0.7934 - 4 n=−3.2066n = -3.2066

  11. Write the concentration in the form 10n10^n.

    C=10−3.2066 g/mLC = 10^{-3.2066} \text{ g/mL}

    The concentration expressed in the form 10n10^n is 10−3.2066\boxed{10^{-3.2066}}.

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